https://adventofcode.com/2021/day/24
I have not completed the day 24 puzzle. The day 24 puzzle has a sequence of instructions representing a calculation to verify a model number (and conceptually enable features based on that model number -- though part A of the puzzle does not provide any details about that). The processing unit performs integer calculations, and has four registers: W, X, Y and Z. There are six instructions, one which inputs a digit of the model number, an add instruction, a multiply instruction, an integer division instruction, a modulo instruction and an equals instruction. The input instruction always inputs to register W (and is the only instruction used to update W). The div instruction always divides by 1 or 26. The modulo instruction is always used to find a remainder modulo 26 (and always only operates on positive values, or 0 for the numerator). The multiply instruction seems to always multiply by 0 or powers of 26. (To load a value into a register, the register is first multiplied by 0 and then has another value added to it.) A certain amount of simplifications are possible using math identities and range constraints. But I do not have much more that's useful to say until I've found a way of solving the puzzle. Here's a snapshot of where I'm at, trying to work through these issues (looks sloppy with a proportionally spaced font): ... digit13 NB. W234=: W216 inp 13 [ 1 thru 9 0 NB. X235=: X223 mul 0 [ 0 add Z228 Y232 NB. X236=: X235 add Z233 [ 9 thru 5520918021 mod 26|Z233 26 NB. X237=: X236 mod 26 [ 0 thru 25 mul <.Z210%26 X223 NB. Z238=: Z233 div 26 [ 0 thru 212343000 add X237 _4 NB. X239=: X237 add -4 [ _4 thru 21 eql X239 digit13 NB. X240=: X239 eql W234 [ 0 thru 1 eql X240 0 NB. X241=: X240 eql 0 [ 0 thru 1 0 NB. Y242=: Y232 mul 0 [ 0 25 NB. Y243=: Y242 add 25 [ 25 mul 25 X241 NB. Y244=: Y243 mul X241 [ 0 thru 25 add Y244 1 NB. Y245=: Y244 add 1 [ 1 thru 26 mul <.Z228%26 Y245 NB. Z246=: Z238 mul Y245 [ 0 thru 5520918000 0 NB. Y247=: Y245 mul 0 [ 0 digit13 NB. Y248=: Y247 add W234 [ 1 thru 9 add digit13 7 NB. Y249=: Y248 add 7 [ 8 thru 16 mul Y249 X241 NB. Y250=: Y249 mul X241 [ 0 thru 16 add Z246 Y250 NB. Z251=: Z246 add Y250 [ 0 thru 5520918016 I am currently working on some tree unification mechanisms (maximum common subtree elimination to let me better inspect the partially resolved calculations). I believe Eugene Nonko has solved this one, though I do not know if he found J useful in his approach. FYI, -- Raul ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
